NIMCET 2023 — Mathematics PYQ

1. Understand the Product Notations

We are given two product terms:

$$\prod _{i=1}^n\tan ({\alpha }_i)=\tan ({\alpha }_1)\cdot \tan ({\alpha }_2)\cdots \tan ({\alpha }_n)=1$$

We want to maximize the target function $P$:

$$P=\prod _{i=1}^n\sin {\alpha }_i=\sin ({\alpha }_1)\cdot \sin ({\alpha }_2)\cdots \sin ({\alpha }_n)$$

2. Relate the Trigonometric Ratios

Let's construct an identity connecting sine, cosine, and tangent. We know that:

$${\sin }^2{\alpha }_i=\frac{{\sin }^2{\alpha }_i}{1}=\frac{{\sin }^2{\alpha }_i}{{\sin }^2{\alpha }_i+{\cos }^2{\alpha }_i}$$

Dividing the numerator and the denominator by ${\cos }^2{\alpha }_i$ gives:

$${\sin }^2{\alpha }_i=\frac{{\tan }^2{\alpha }_i}{{\tan }^2{\alpha }_i+1}$$

Taking the product from $i=1$ to $n$ on both sides:

$$\prod _{i=1}^n{\sin }^2{\alpha }_i=\prod _{i=1}^n\frac{{\tan }^2{\alpha }_i}{1+{\tan }^2{\alpha }_i}$$

Since ${\prod }_{i=1}^n\tan ({\alpha }_i)=1$, squaring it means ${\prod }_{i=1}^n{\tan }^2{\alpha }_i=1$. The expression simplifies to:

$${\left(\prod _{i=1}^n\sin {\alpha }_i\right)}^2=\frac{1}{{\prod }_{i=1}^n(1+{\tan }^2{\alpha }_i)}$$

3. Apply AM-GM Inequality for Optimization

To maximize the fraction, we need to minimize the denominator term ${\prod }_{i=1}^n(1+{\tan }^2{\alpha }_i)$.

Using the Arithmetic Mean - Geometric Mean (AM-GM) inequality ($a+b\ge 2\sqrt{ab}$) for each term inside the product where $a=1$ and $b={\tan }^2{\alpha }_i$:

$$1+{\tan }^2{\alpha }_i\ge 2\sqrt{1\cdot {\tan }^2{\alpha }_i}$$

$$1+{\tan }^2{\alpha }_i\ge 2\tan {\alpha }_i$$

Taking the product of this inequality from $i=1$ to $n$:

$$\prod _{i=1}^n(1+{\tan }^2{\alpha }_i)\ge \prod _{i=1}^n(2\tan {\alpha }_i)$$

$$\prod _{i=1}^n(1+{\tan }^2{\alpha }_i)\ge 2^n\cdot \prod _{i=1}^n\tan {\alpha }_i$$

Substitute the given value ${\prod }_{i=1}^n\tan {\alpha }_i=1$:

$$\prod _{i=1}^n(1+{\tan }^2{\alpha }_i)\ge 2^n$$

The minimum value of the denominator is $2^n$, which occurs under symmetric conditions when $\tan {\alpha }_1=\tan {\alpha }_2=\cdots =\tan {\alpha }_n=1$ (meaning ${\alpha }_i=\frac{\pi }{4}$).

4. Calculate the Maximum Value

Substitute the minimum value of the denominator back into our equation from Step 2:

$${\left(\prod _{i=1}^n\sin {\alpha }_i\right)}^2\le \frac{1}{2^n}$$

Taking the square root on both sides to find the maximum value:

$$\prod _{i=1}^n\sin {\alpha }_i\le \sqrt{\frac{1}{2^n}}$$

$$\prod _{i=1}^n\sin {\alpha }_i\le \frac{1}{(2^n{)}^{1/2}}$$

$$\prod _{i=1}^n\sin {\alpha }_i\le \frac{1}{2^{n/2}}$$

Therefore, the maximum value is $\frac{1}{2^{n/2}}$.